package com.example.demo.leetcode.top;

/**
 * 给你一个由 '1'（陆地）和 '0'（水）组成的的二维网格，请你计算网格中岛屿的数量。
 *
 * 岛屿总是被水包围，并且每座岛屿只能由水平方向和/或竖直方向上相邻的陆地连接形成。
 *
 * 此外，你可以假设该网格的四条边均被水包围。
 *
 *  
 *
 * 示例 1：
 *
 * 输入：grid = [
 *   ["1","1","1","1","0"],
 *   ["1","1","0","1","0"],
 *   ["1","1","0","0","0"],
 *   ["0","0","0","0","0"]
 * ]
 * 输出：1
 * 示例 2：
 *
 * 输入：grid = [
 *   ["1","1","0","0","0"],
 *   ["1","1","0","0","0"],
 *   ["0","0","1","0","0"],
 *   ["0","0","0","1","1"]
 * ]
 * 输出：3
 *  
 *
 * 提示：
 *
 * m == grid.length
 * n == grid[i].length
 * 1 <= m, n <= 300
 * grid[i][j] 的值为 '0' 或 '1'
 *
 * 来源：力扣（LeetCode）
 * 链接：https://leetcode-cn.com/problems/number-of-islands
 * 著作权归领扣网络所有。商业转载请联系官方授权，非商业转载请注明出处。
 **/
public class Test_岛屿数量 {

    public int numIslands(char[][] grid) {

        int count=0;
        for(int x=0;x<grid.length;x++){
            for(int y=0;y<grid[0].length;y++){
                if(grid[x][y]=='1'){
                    dfsTrunToZERO(grid,x,y);
                    count++;
                }
            }
        }
        return count;
    }

    /**
     * 将可以遍历到的陆地全部置为2
     * @param grid
     * @param x
     * @param y
     */
    public void dfsTrunToZERO(char[][] grid,int x,int y){
        if(isExists(grid,x,y)){
            if(grid[x][y]=='2'){
                return;
            }
            grid[x][y]='2'; //遍历过了
            dfsTrunToZERO(grid,x+1,y);
            dfsTrunToZERO(grid,x,y+1);
            dfsTrunToZERO(grid,x-1,y);
            dfsTrunToZERO(grid,x,y-1);
        }
    }


    /**
     * 判断当前坐标点，是否存在数组内,并且是1
     * @param grid
     * @param x
     * @param y
     * @return
     */
    public boolean isExists(char[][] grid,int x,int y){
        if(x>=0 && x<grid.length && y>=0 && y<grid[0].length && grid[x][y]=='1'){
            return true;
        }
        return false;
    }


    public static void main(String[] args) {

    }
}
